CUET UG Physics Booster Test 2-Diode Applications and Circuits -

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CUET UG Physics Booster Test 2-Diode Applications and Circuits

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2-Diode Applications and Circuits

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QUESTION 1 OF 20

Incorrect statement about metallic contacts in a p-n junction diode

Answer Review

1 Incorrect statement about metallic contacts in a p-n junction diode

A They are provided at the ends of the p-n junction.

B They are necessary for connecting the diode to an external circuit.

C They are responsible for generating the depletion region.

D They allow the application of an external bias voltage.

💡 Short Explanation

Metallic contacts provide electrical connections. The depletion region is formed due to diffusion and recombination of charge carriers.

✅ Correct Answer Explanation

The depletion region forms naturally at the p-n junction because electrons and holes diffuse across the junction and recombine. Metallic contacts merely allow connection to external circuits.

❓ Why Other Options Are Incorrect

A: Correct.
B: Correct.
D: Correct.

🎯 Strategy

Used

Conceptual Elimination

Final Logic:

Contacts do not create the depletion region.

🧠 Memory Trick

"Contacts Connect, Junction Creates Depletion."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: p-n Junction Diode

2 Diode symbol statements
1. It is a two-terminal device.
2. The arrow points in the direction of conventional current under forward bias.
3. The arrow represents the physical movement of electrons from n-side to p-side.
4. The symbol simplifies the representation of the p-n junction in circuit diagrams.

A (A), (B), (D)

B (A), (C), (D)

C (B), (C), (D)

D (A), (B), (C)

💡 Short Explanation

Diode is a two-terminal device. Symbol shows conventional current direction. It simplifies circuit representation.

✅ Correct Answer Explanation

Statements A, B, and D are correct. The arrow indicates conventional current flow, not electron flow.

❓ Why Other Options Are Incorrect

C: Arrow does not represent electron motion.

🎯 Strategy

Used

Symbol Interpretation

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"Arrow = Conventional Current, Not Electrons."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Diode Symbol

3 Correct statements about the variation of current with voltage
1. In forward bias, the current is measured in mA and depends on minority carrier injection.
2. In reverse bias, the drift current is measured in μA and is driven by the electric field sweeping minority carriers.
3. The forward bias resistance is exceptionally high compared to the reverse bias resistance.
4. The reverse current is essentially independent of voltage up to the breakdown limit.

A (A), (B), (C)

B (A), (B), (D)

C (B), (C), (D)

D (A), (C), (D)

💡 Short Explanation

Forward current is usually measured in mA. Reverse current is usually measured in μA. Reverse current remains nearly constant before breakdown.

✅ Correct Answer Explanation

Forward conduction involves carrier injection and large currents, whereas reverse current is due to minority carriers and remains almost constant up to breakdown.

❓ Why Other Options Are Incorrect

C: Forward resistance is much lower than reverse resistance.

🎯 Strategy

Used

V-I Characteristics Analysis

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"Forward: Low Resistance, Reverse: High Resistance."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: V-I Characteristics of p-n Junction Diode

4 Match List I (Circuit Condition) with List II (Expected Value/Measurement)

List I	List II
1. Forward bias applied voltage	i. Measured in microamperes (μA)
2. Forward bias current	ii. Measured in volts (e.g., 0 to 1 V)
3. Reverse bias applied voltage	iii. Measured in milliamperes (mA)
4. Reverse bias current	iv. Measured in tens of volts (e.g., 0 to −10 V)

A a–ii, b–iii, c–iv, d–i

B a–iii, b–ii, c–i, d–iv

C a–iv, b–i, c–ii, d–iii

D a–i, b–iv, c–iii, d–ii

💡 Short Explanation

Forward bias voltage is usually small and measured in volts. Forward bias current is comparatively large and measured in milliamperes. Reverse bias voltage is often applied in a wider voltage range. Reverse current is very small and measured in microamperes.

✅ Correct Answer Explanation

In a p–n junction diode, the forward bias voltage required for conduction is typically less than 1 V, so it is measured in volts. Hence, Forward bias applied voltage → ii. The resulting forward current is relatively large and lies in the milliampere range, giving Forward bias current → iii. Under reverse bias, a larger voltage is generally applied across the diode, often extending to several volts or tens of volts. Therefore, Reverse bias applied voltage → iv. The reverse current remains extremely small and is measured in microamperes, giving Reverse bias current → i. Thus, the correct matching is: a–ii, b–iii, c–iv, d–i

❓ Why Other Options Are Incorrect

Option B → a–iii, b–ii, c–i, d–iv

Voltage and current units are interchanged in both forward and reverse bias conditions.

Option C → a–iv, b–i, c–ii, d–iii

Forward bias quantities are incorrectly assigned reverse-bias measurement ranges.

Option D → a–i, b–iv, c–iii, d–ii

None of the voltage-current pairings correspond to actual diode characteristics.

🎯 Strategy

Used

Quantity-and-Unit Matching

Application:

Recall the typical magnitude of diode voltages and currents:
Forward voltage → Volts
Forward current → mA
Reverse voltage → Tens of volts
Reverse current → μA

Final Logic:

Match each diode parameter with its commonly used measurement unit and range.

🧠 Memory Trick

Reverse bias → Larger voltage, tiny current

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits – Topic: V–I Characteristics of a p–n Junction Diode

5 When the applied forward voltage crosses the threshold voltage:

A the depletion region width significantly increases.

B the current begins to increase exponentially even for a very small increase in bias voltage.

C the barrier height reaches its absolute maximum value.

D the diode immediately enters the reverse breakdown region.

💡 Short Explanation

Threshold voltage marks rapid increase in current.

✅ Correct Answer Explanation

Once the cut-in voltage is reached, a small increase in forward voltage causes a large increase in current.

❓ Why Other Options Are Incorrect

A: Depletion width decreases.
C: Barrier height decreases.
D: Reverse breakdown is unrelated.

🎯 Strategy

Used

Forward-Bias Analysis

Final Logic:

Current rises exponentially beyond threshold voltage.

🧠 Memory Trick

"Cross Threshold → Current Shoots Up."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Threshold Voltage

6 The typical cut-in voltage for a silicon diode is roughly:

A 2.0 × 10⁻¹ V

B 3.0 × 10⁰ V

C 1.1 × 10⁰ V

D 7.0 × 10⁻¹ V

💡 Short Explanation

Silicon diode cut-in voltage ≈ 0.7 V.

✅ Correct Answer Explanation

\(V_{cut-in}\approx 0.7V\) for a silicon diode.

❓ Why Other Options Are Incorrect

A: Corresponds approximately to germanium diode.
B, C: Too large.

🎯 Strategy

Used

Fact Recall

Final Logic:

Silicon diode cut-in voltage is approximately 0.7 V.

🧠 Memory Trick

"Si = 0.7 V, Ge = 0.2 V."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Cut-in Voltage

7 The reverse saturation current of a diode is:

A heavily dependent on the magnitude of the applied reverse voltage.

B limited by the concentration of minority carriers on either side of the junction.

C caused primarily by the diffusion of majority carriers across the depletion layer.

D usually in the range of several milliamperes.

💡 Short Explanation

Reverse saturation current is caused by minority carriers.

✅ Correct Answer Explanation

Minority carriers swept across the junction by the electric field produce the reverse saturation current.

❓ Why Other Options Are Incorrect

A: Nearly independent of reverse voltage before breakdown.
C: Not caused by majority carriers.
D: Usually in μA range.

🎯 Strategy

Used

Current Mechanism Analysis

Final Logic:

Minority carrier concentration limits reverse current.

🧠 Memory Trick

"Reverse Current = Minority Carrier Current."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Reverse Saturation Current

8 If the reverse current is not limited by an external circuit and exceeds the rated value at the breakdown voltage, the p-n junction will be:

A Destroyed, Overheating

B Cooled, Preserved

C Saturated, Rectification

D Charged, Capacitance

💡 Short Explanation

Excessive breakdown current generates heat.

✅ Correct Answer Explanation

If current is not limited, excessive power dissipation causes overheating and permanent damage to the junction.

❓ Why Other Options Are Incorrect

They do not represent the actual consequence of excessive breakdown current.

🎯 Strategy

Used

Breakdown Analysis

Final Logic:

Uncontrolled breakdown can destroy the diode.

🧠 Memory Trick

"Breakdown Without Limiting = Damage."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Breakdown Voltage

9 Dynamic resistance expressions
1. It is the inverse of the slope of the V-I characteristic curve.
2. It is constant across all voltages in a p-n junction.
3. It is lower in forward bias than in reverse bias.
4. It is defined as the ratio of a small change in voltage to a small change in current.

A (A), (B), (C)

B (A), (B), (D)

C (A), (C), (D)

D (B), (C), (D)

💡 Short Explanation

Dynamic resistance is not constant. It equals ΔV/ΔI.

✅ Correct Answer Explanation

\(r_{d}=\frac{\Delta V}{\Delta I}\) It is the inverse of the slope of the I-V curve and is much lower in forward bias.

❓ Why Other Options Are Incorrect

B: Dynamic resistance changes with operating point.

🎯 Strategy

Used

Formula Analysis

Final Logic:

A, C, and D are correct.

🧠 Memory Trick

"Dynamic Resistance Changes with Bias."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Dynamic Resistance

10 The dynamic resistance calculated at a reverse bias voltage of V = −10 V (where I = −1 μA) is:

A 1.0 × 10¹ Ω

B 1.0 × 10³ Ω

C 1.0 × 10⁵ Ω

D 1.0 × 10⁷ Ω

💡 Short Explanation

Reverse bias dynamic resistance is extremely high.

✅ Correct Answer Explanation

From the reverse V-I characteristic discussed in NCERT Example 14.4, the calculated dynamic resistance is: \(r_{d}=1.0\times {10}^{7}\Omega\) showing the very high resistance offered in reverse bias.

❓ Why Other Options Are Incorrect

They are much smaller than the experimentally determined reverse dynamic resistance.

🎯 Strategy

Used

NCERT Example Recall

Final Logic:

Reverse dynamic resistance ≈ \({10}^{7}\Omega\).

🧠 Memory Trick

"Reverse Bias → Very High Resistance."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Dynamic Resistance from V-I Characteristics

11 Incorrect statement about unidirectional flow

A A p-n junction diode offers low resistance when forward biased.

B A p-n junction diode offers high resistance when reverse biased.

C This property of variable resistance prevents the rectification of AC voltages.

D Current flows mainly in the part of the cycle where the diode is forward biased.

💡 Short Explanation

The diode's unidirectional property enables rectification. It does not prevent rectification.

✅ Correct Answer Explanation

A p-n junction diode offers low resistance in forward bias and high resistance in reverse bias. This behavior is precisely what makes AC-to-DC conversion possible.

❓ Why Other Options Are Incorrect

A: Correct statement.
B: Correct statement.
D: Correct statement.

🎯 Strategy

Used

Conceptual Elimination

Final Logic:

Rectification occurs because of unidirectional conduction.

🧠 Memory Trick

"One-Way Conduction = Rectification."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Rectification

12 Correct statements about AC transformation using diodes
1. The process is known as rectification.
2. It utilizes the unidirectional conducting property of a p-n junction.
3. The secondary of a transformer supplies the desired AC voltage.
4. The output of a basic rectifier circuit is a pure, ripple-free DC voltage without filters.

A (A), (B), (C)

B (A), (C), (D)

C (B), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Rectification converts AC into DC. Transformer secondary provides AC input. Filters are required for ripple reduction.

✅ Correct Answer Explanation

Statements A, B, and C correctly describe rectification. A basic rectifier produces pulsating DC, not pure DC.

❓ Why Other Options Are Incorrect

D: Incorrect because filters are needed to obtain smoother DC.

🎯 Strategy

Used

Concept Identification

Final Logic:

Only A, B, and C are correct.

🧠 Memory Trick

"Rectifier Converts, Filter Smooths."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Rectifiers

13 Match List I (Circuit Event) with List II (Component/Voltage State) for a half-wave rectifier

List I	List II
1. Terminal A is positive	i. During negative half-cycle
2. Terminal A is negative	ii. Diode conducts
3. Output voltage appears	iii. Diode does not conduct
4. Diode is reverse-biased	iv. During positive half-cycle

A a–ii, b–i, c–iv, d–iii

B a–iii, b–ii, c–i, d–iv

C a–iv, b–iii, c–ii, d–i

D a–i, b–iv, c–iii, d–ii

💡 Short Explanation

A positive terminal A forward-biases the diode. A negative terminal A corresponds to the negative half-cycle. Output appears during the conducting half-cycle. A reverse-biased diode blocks current.

✅ Correct Answer Explanation

When Terminal A is positive, the diode becomes forward-biased and conducts current. Therefore, a → ii (Diode conducts). When Terminal A is negative, it corresponds to the negative half-cycle, giving b → i. In a half-wave rectifier, output voltage appears only during the positive half-cycle when the diode conducts, so c → iv. A reverse-biased diode prevents current flow and therefore does not conduct, giving d → iii. Thus, the correct matching is: a–ii, b–i, c–iv, d–iii

❓ Why Other Options Are Incorrect

Option B → a–iii, b–ii, c–i, d–iv

A positive terminal does not make the diode non-conducting, and output does not appear during the negative half-cycle.

Option C → a–iv, b–iii, c–ii, d–i

Terminal polarity is incorrectly associated with half-cycles and conduction states.

Option D → a–i, b–iv, c–iii, d–ii

Reverse-biased diodes do not conduct, making this matching incorrect.

🎯 Strategy

Used

Half-Wave Rectifier Logic

Application:

Determine diode behavior during positive and negative half-cycles.

Final Logic:

Positive terminal → Diode conducts
Negative terminal → Negative half-cycle
Output voltage → Positive half-cycle
Reverse bias → No conduction

🧠 Memory Trick

Output → Positive Half-Cycle

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics: Materials, Devices and Simple Circuits – Topic: Half-Wave Rectifier

14 The rectified output of a half-wave rectifier contains voltage restricted to one direction, and it mathematically represents:

A Both halves of the AC wave

B Half of the input AC wave

C Double the frequency of the input AC wave

D A pure, steady DC voltage

💡 Short Explanation

Only one half-cycle is utilized.

✅ Correct Answer Explanation

A half-wave rectifier passes only one half of the AC input waveform while blocking the other half.

❓ Why Other Options Are Incorrect

A: Both halves occur only in full-wave rectification.
C: Frequency doubling occurs in full-wave rectification.
D: Output is pulsating DC.

🎯 Strategy

Used

Waveform Analysis

Final Logic:

Half-wave rectifier outputs half of the AC waveform.

🧠 Memory Trick

"Half-Wave = Half the Wave."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Half-Wave Rectifier Output

15 In a full-wave rectifier, because the centre-tap provides the reference ground, the voltage rectified by each diode at any instant is:

A twice the total secondary voltage.

B equal to the peak secondary voltage of the entire coil.

C zero, as it is perfectly cancelled out.

D only half the total secondary voltage.

💡 Short Explanation

Each diode uses one half of the secondary winding.

✅ Correct Answer Explanation

The centre tap divides the secondary winding into two equal halves. Each diode rectifies the voltage across one half of the winding.

❓ Why Other Options Are Incorrect

A, B: Overestimate the voltage.
C: Voltage is not zero.

🎯 Strategy

Used

Transformer Analysis

Final Logic:

Each diode receives half of the secondary voltage.

🧠 Memory Trick

"Centre Tap Splits the Secondary."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Full-Wave Rectifier

16 In a full-wave rectifier, during the positive half cycle, Diode D1 is ________ and during the negative half cycle, Diode D2 is ________.

A Forward-biased, Forward-biased

B Reverse-biased, Reverse-biased

C Forward-biased, Reverse-biased

D Reverse-biased, Forward-biased

💡 Short Explanation

D1 conducts during positive half-cycle. D2 conducts during negative half-cycle.

✅ Correct Answer Explanation

Each diode becomes forward biased during its respective half-cycle, ensuring current through the load remains in the same direction.

❓ Why Other Options Are Incorrect

They do not match actual full-wave rectifier operation.

🎯 Strategy

Used

Diode Conduction Analysis

Final Logic:

Both diodes conduct alternately in forward bias.

🧠 Memory Trick

"Positive → D1 ON, Negative → D2 ON."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Full-Wave Rectifier Operation

17 If no external load is connected to a full-wave rectifier equipped with a capacitor input filter:

A the capacitor immediately discharges to zero volts.

B the capacitor remains charged to the peak voltage of the rectified output.

C the capacitor reverses its polarity periodically.

D the capacitor acts as a short circuit and destroys the diodes.

💡 Short Explanation

Without a load, there is no discharge path.

✅ Correct Answer Explanation

The capacitor charges to the peak output voltage and remains charged because no load is available to discharge it.

❓ Why Other Options Are Incorrect

A: No discharge path exists.
C: Capacitor polarity does not reverse.
D: Capacitor does not short-circuit the circuit.

🎯 Strategy

Used

Filter Analysis

Final Logic:

No load means no significant discharge.

🧠 Memory Trick

"No Load = Charge Held."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Capacitor Input Filter

18 Inductor series statements
1. An inductor can be used in series with the load.
2. It serves as a filter.
3. It helps to filter out the AC ripple.
4. It increases the pulsating nature of the output voltage.

A (A), (C), (D)

B (A), (B), (D)

C (B), (C), (D)

D (A), (B), (C)

💡 Short Explanation

Inductors oppose current fluctuations. They reduce ripple.

✅ Correct Answer Explanation

An inductor placed in series with the load acts as a filter and smooths current variations, thereby reducing ripple.

❓ Why Other Options Are Incorrect

D statement: Ripple is reduced, not increased.

🎯 Strategy

Used

Filter Function Analysis

Final Logic:

A, B, and C are correct.

🧠 Memory Trick

"Inductor Smooths Current."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Inductor Filter

19 Incorrect statement about time constants

A The time constant is the product of capacitance C and effective resistance Rₗ.

B A large time constant ensures a slow rate of voltage fall across the capacitor.

C To make the time constant large, the value of the capacitor should be very small.

D Capacitor input filters commonly use large capacitors to maintain stability.

💡 Short Explanation

The time constant is: \(\tau =R_{L}C\)

✅ Correct Answer Explanation

A larger capacitance increases the time constant, resulting in slower discharge and reduced ripple.

❓ Why Other Options Are Incorrect

A, B, D: Correct statements.

🎯 Strategy

Used

Formula Application

Final Logic:

Large capacitance → large time constant.

🧠 Memory Trick

"Big C = Big Time Constant."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Capacitor Filters

20 Ripple removal concepts
1. The rate of fall of voltage across the discharging capacitor dictates the ripple size.
2. A larger capacitance effectively minimizes ripple.
3. The output voltage is perfectly constant DC regardless of load resistance.
4. The charging and discharging cycles of the capacitor form the remaining ripple.

A (A), (B), (C)

B (A), (B), (D)

C (A), (C), (D)

D (B), (C), (D)

💡 Short Explanation

Ripple depends on capacitor charging and discharging. Larger capacitance reduces ripple.

✅ Correct Answer Explanation

Ripple is produced due to the periodic charging and discharging of the capacitor through the load. Increasing capacitance increases the time constant and decreases ripple.

❓ Why Other Options Are Incorrect

C: Output is never perfectly constant DC in practical rectifier circuits.

🎯 Strategy

Used

Filter Performance Analysis

Final Logic:

A, B, and D are correct.

🧠 Memory Trick

"Large Capacitor → Small Ripple."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Ripple Removal and Filters

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